CBSE Class 9 Science Structure Of Atom Notes

STRUCTURE OF ATOM

INTRODUCTION

(i) The idea of tiniest unit of matter (Anu and Parmanu) was propounded by maharishi Kanad in Vedic period in our country.

(ii) Democritus, a Greek philosopher also proposed that matter is made up of extremely small particles, the  “atom”. The name atom comes from Greek language.

(iii) John Dalton in 1808 published theory of atom assuming that atoms are the ultimate indivisible particles of matter.

(iv) Later the works f William Crookes (1878) , J.J. Thomson (1897) and Goldstein proved that atom of any element contains smaller particles which are either positively charged or negatively charged.

(v) Work of Rutherford and Neils Bohr confirmed that an atom consists of three subatomic particles, that are electrons, protons and neutrons.

(vi) It has been established that the central core of an atom consists of protons and neutrons and is commonly called nucleus. The electrons revolve around the nucleus.

(vii) The atom as a whole is electrically neutral as the number of protons in it, is equal to the number of electrons.

The smallest indivisible particle or unit of an element is called an atom, which can take part in a chemical reaction and may or may not exist independently.

An element is a pure substance which cannot be subdivided into two or more new substances by any means.

The word “ATOM” is given by “John Dalton”.

FUNDAMENTAL PARTICLES OF AN ATOM

(a) Electron :

Electron has a negative charge on it, its mass is 1/1837 times the mass of one atom of hydrogen. It is denoted by the symbol  ⁰ _-1 θ, where 0 denotes its mass and -1 denotes its charge. Electrons in the outer most shell are called valence electrons.

(b) Proton :

Proton has a unit positive charge, it is denoted by the symbol ¹₊₁ p, where 1 denotes it atomic mass and +1 denotes its charge.

(c) Neutron :

Neutron has no electric charge on it. Its mass is almost equal to the mass of one atom of hydrogen. it is denoted by the symbol ¹₀ n, where 1 denotes its atomic mass and 0 denotes its charge.

In the neutral atom the total number of protons in the nucleus is equal to the number of electrons revolving round the nucleus.

THOMSONS MODEL OF AN ATOM

After the discovery of electrons and protons J.J. Thomson (1898) tried to explain the arrangement of electrons and protons within the tom. He proposed that an atom consists of a sphere of positive electricity in which electrons are embedded like plum in pudding or seeds evenly distributed in red spongy mass in watermelon. The radius of the sphere is of the other 10^-8 cm which is equal to the size of the atom. Although Thmoson’s model could explain the electrical neutrality of an atom but this model could not satisfy experimental facts proposed by Rutherford and hence was discarded.

 “Proton is a sub - atomic particle having a unit positive charge (+1.602 × 10^-19 kg) & mass (1.6725 × 10^-27 kg) which is about 1837 times greater than the mass of an electron.”

RUTHERFORD’S MODEL OF AN ATOM

 (a) Rutherford’s Alpha Scattering Experiment :

Ernest Rutherford and his coworkers performed numerous experiments in which  particles emitted from a radioactive element such as polonium were allowed to strike thin sheets of metals such as gold and platinum.

(i) A beam of -particles (He²⁺) we obtained by placing polonium in a lead box and letting the alpha particles come out of a pinhole in the lead box. This beam of -rays was directed against a thin gold foil (0.00004 cm). A circular screen coated with zinc sulphide was placed on the other side of the foil.

(ii) About 99% of the  particles passed undeflected through the gold foil and caused illumination of zinc sulphide screen.

(iii) Very few -particles underwent small and large deflections after passing through the gold foil.

(iv) A very few (about 1 in 20,000) were deflected backward on their path at an angle of 1800.

Rutherford was able to explain these observations as follows :

(i) Since a large number of -particles pass through the atom undeflected, hence, there must be large empty space within the atom.

(ii) As some of the - particles got defleted, therefore, there must be something massive and positively charged present in the atom.

(iii) The number of -particles which got deflected is very small, therefore, the whole positive charge in the atom is conventratted in a very small space.

(iv) Some of the -particles retracted their path i.e. came almost straight back towards the sources as a result of their direct collisions with the heavy mass.

- particles are made up of two protons and two neutrons and are He nuclei.

The positively charged heavy mass which occupies only a small space as compared to the total space occupied by the atom is called nucleus.

(b) Rutherford’s Nuclear Model of Atom :

Rutherford proposed a new picture of the structure of the atom. Main feature of this model are as follows -

(i) The atom of an element consists of a small positively charged “Nucleus” which is situated at the centre of the atom and which carries almost the entire mss of the atom.

(ii) The electrons are distributed in the empty space of the atom around the nucleus in different concentric circular paths (orbits).

(iii) The number of electrons in the orbits is equal to the total number of protons in the nucleus.

(iv) Volume of nucleus is very small as compared to the volume of atom.

(v) Most of the space in the atom is empty.

The stability of such a system in which negatively charged electrons surround a positively charged nucleus was explained by proposing that the electrons revolve around the nucleus at very high speed in circular orbits. This arrangement is just like our solar system. The high sped of the moving electrons given them a centrifugal force acting away from the nucleus. The centrifugal force balance the electrostatic force of attraction acting between the nucleus and the electrons.

(c) Defects in Rutherford’s Model :

(i) Rutherford did not specify the number of electrons in each orbit.

(ii) According to electromagnetic theory, if a charged particle (electron) is accelerated around another charged particle (nucleus) then there would be continuous radiation of energy. This loss of energy would slow down the  speed of electron and eventually the electron would fall into the nucleus. But such a collapse does not occur. Rutherford’s model could not explain this theory.

BOHR’S MODEL OF AN ATOM

Rutherford’s model of the atom was unable to explain certain observations with regard to the atom i.e. stability of the atom and the occurrence of the atomic spectra. Neils Bohr accepted Rutherford’s idea that the positive charge and most of the mass of the atom is concentrated in its nucleus with the electrons present at some distance away.

According to Bohr’s theory -

(i) Electrons revolve around the nucleus in well defined orbits or shells, each shell having a definite amount of energy associated with the electrons in it. Therefore, these shells are also called energy levels.

(ii) The energy associated with the electrons in an orbit increases as the radius of the orbit increases. These shells are known as K, L, M, N etc. starting from the one closest to the nucleus. 

(iii) An electron in a shell can more to a higher or lower energy shell by absorbing or releasing a fixed amount of energy. 

(iv) The amount of energy absorbed or emitted is given by the difference of energies associated with the two energy levels.       

Energy absorbed,        DE = E2 - E1 =  hν       

Energy emitted,           DE = E2 - E1 = hν 

Where h is Plank’s constant (h = 6.62 × 10^-34 Js) and  is the frequency of the radiation.

ATOMIC STRUCTURE

An atom consists of two parts :

(a) Nucleus :

Nucleus is situated in the centre of an atom. All the protons & neutrons are situated in the nucleus, therefore, the entire mass of an atom is almost concentrated in the nucleus. The overall charge of nucleus is positive due to the presence of positively charged protons (neutrons present have no charge). The protons & neutrons are collectively called nucleons. The radius of the nucleus of an atom is of the order of 10^-15 m.

(b) Extra nuclear region :

In extra nuclear part or in the region outside the nucleus, electrons are present which revolve around the nucleus in orbits of fixed energies. These orbits are called energy levels. These energy levels are designated as K, L, M, N and so on.

The maximum number of electrons that can be accommodated in a shell is given by the formula 2n² (where n = number of shell i.e. 1, 2, 3--)

Each energy level is further divided into subshells designated as s.p.d.f.

1st shell (K) contains 1 subshell (s)

2nd shell (L) contains 2 subshells (s,p)

3rd shell ( M) contains 3 subshells (s,p,d)

4th shell (N) contains 4 subshells (s,p,d,f).

ORBITALS

Like shells are divided into subshells, subshells further contain orbitals.

An orbital may be defined as a 

“Region in the three - dimensional space around the nucleus where the probability of finding an electron is maximum. The maximum capacity of each orbital is that of two electrons.”

The total number of nucleons is equal to the mass number (A) of the atom.

ELECTRONIC CONFIGUATION OF AN ATOM

(i) Each of the orbits can accommodate a fixed number of electrons. Maximum number of electrons in an orbit is equal to 2n², where ‘n’ is the number of the orbit.

If n = 1 then 2n2 = 2

 n = 2 then 2n2 = 8

n = 3 then 2n2 = 18

n = 4 then 2n2 = 32

(ii) In the outermost shell of any atom, the maximum possible number of electrons is 8, except in the first shell which can have at the most 2 electrons.

(iii) The arrangement of the electrons is different shells is known as the electronic configuration of the element. The pictorial representations of Bohr’s model o hydrogen, helium, carbon, sodium and calcium atoms having 1, 2, 6, 11 and 20 electrons respectively are shown in the figure where the centre of the circle represents the nucleus.

If the outermost shell has 8 electrons it is said to be an octet. If the first shell has its full quota of 2 electrons, it is said to be duplet.

ELECTRONIC CONFIGURATION OF ELEMENTS UPTO ATOMIC NUMBER 30

VALENCY

Valency of an element is the combining capacity of the atoms of the element with atoms of the same or different elements. The combining capacity of the atoms of other elements was explained in terms of their tendency to attain a fully - filled outermost shell (stable octet or duplet).

The number of electrons gained, lost or contributed for sharing by an atom of the element gives us directly the combining capacity or valency of the element.

Valency of an element is determined by the number of valence electrons in an atom of the element.

The valency of an element        =          number of valence electrons 

                                                                 (when number of valence electrons are from 1 to 4)

The valency of an element        =          8 - number of valence electrons.

                                                                   (when number of valence electrons are more than 4)

Na has 1 valence electron, thus, its valency is 1.

Cl has 7 valence electrons, thus, its valency is 8 - 7 = 1.

ATOMIC NUMBER (Z)

The number of protons is the nucleus of an atom of a given element is called the atomic number of that element.

                                                             or        

Atomic number is the number of protons present in the atom of an element. It is denoted. by “Z”

Atomic number = Number of protons = Number of electrons ( in a neutral atom)

 Atomic number = Number of protons    (in an ion)

e.g. ₁₁Na

-Atomic number of sodium is 11

 -Nucleus of sodium has 11 protons.

- Nucleus of sodium has 11 units of positive charge.

- There are 11 electrons, revolving round the nucleus of sodium.

The atomic number is represented on the LHS of the symbol of the element as subscript.

MASS NUMBER (A)

Mass number is the number of protons and neutrons present in the atom of an element. It is denoted by “A”. The mass number is represented either on the left hand side (LHS) or on the right hand side (RHS) of the symbol of the element as superscript.

Mass number = Number of protons + Number of neutrons.

e.g.

-Mass number of aluminum is 27.

-The total number of protons and neutrons is the nucleus of aluminum is 27.

-Number of protons is 13.

-Number of neutrons is = 27 - 13 = 14.

Each element has a unique atomic number which is its identity.

(a) Relation between Z, A and N

A = Z + N

Z = Number of Protons

N = Number of neutrons

A = Mass number

N = A - Z

ISOTOPES 

(a) Atoms of same element having the same chemical properties, but differing in mass are known as isotopes.

The isotopes of an element have the same atomic number but different atomic masses. Isotopes have the same electrical charges means same number of protons. The difference in their masses is due to the presence of different number of neutrons.

e.g. (a) Isotopes of hydrogen

All the isotopes of an element have identical chemical properties.

(b) Characteristics of Isotopes :

(i) The physical properties of the isotopes of an element are different number of neutrons in their nuclei. Hence mass, density and other physical properties of the isotopes of an element are different.

(ii) All the isotopes of an element contains the same number of electrons. So, they have the same electronic configuration with the same number of valence electrons. Since the chemical properties of an element are determined by the number of valence electrons in its atom, all the isotopes of an element have identical chemical properties.

(c) Reason for the fractional atomic masses of elements :

The atomic masses of many elements are in fraction and not whole number. The fractional atomic masses of elements are due to the existence of their isotopes having different masses.

e.g. :The atomic mass of chlorine is 35.5 u. Chlorine has two isotopes    with abundance of 75% and 25% respectively. Thus the average mass of a chlorine atom will be 75% of Cl - 35 and 25% of Cl-37, which is 35.5 u.

i.e., average atomic mass of chlorine

Thus, the average atomic mass of chlorine is 35.5 u.

Similarly, average atomic mass of copper is 63.5 u.

(d) Applications of Radioactive isotopes :

(i) In agriculture : Certain elements such as boron, cobalt, copper, manganese, zinc and molybdenum are necessary is very minute quantities for plant nutrition.

By radioactive isotopes we van identify the presence and requirements of these element in the nutrition of plants.

(ii) In industry : Coating on the arm of clock to seen in dark. To identify the cracks in metal casting.     

(iii) In medicine : Thyroid, bone diseases, brain tumours and cancer and diagnosed, controlled or destroyed with the help of radioactive isotopes like  ,Na, iodine, phosphorus etc.

(iv) Determination of the mechanism of chemical reaction : by replacing an atom or molecule by its isotope.

(v) In carbon dating : Will and Libby (1960) developed the technique of radiocarbon dating to determine the age of plants, fossils and archeological samples.

Isotopes (Like Uranium - 238) are used in nuclear reactor to produce energy and power.

ISOBARS

the atoms of different elements with different atomic numbers, but same mass number are called isobars.

ISOTONES

The isotones may be defined as the atoms of different elements containing same number of neutrons.

ISOELECTRIC

Ion or atom or molecule which have the same number of electrons are called as isoelectronic species.

e.g.                              Cl^-                    Ar                     K⁺                     Ca⁺²

No. of electrons         18                     18                     18                     18

Isobars contain different number of electrons, protons and neutrons

SOLVED EXAMPLES

1Calculate the number of electrons, protons and neutrons in the following species.

 (i) Phosphorus atom      (ii) Phosphide ion (P^3-)   (iii) Magnesium ion (Mg²⁺)

Mass number :             P = 31, Mg = 24

Atomic numbers :          P = 15, Mg = 12

Sol. (i) Phosphorus atom

Number of electrons = Atomic number = 15

Number of protons = Atomic number = 15

Number of neutrons = Mass number - Atomic number = 31 - 15 = 13.

(ii) Phosphide ion (P^3-).

Phosphide ion (P^3-) = Phosphorus atom + 3 electrons

P^3-  P + 3e ^-

Thus, phosphide ion has same number of protons and neutrons as phosphorus atom but has three electrons more.

Number of electrons                  =          15 + 3 = 18

Number of protons                    =          15

Number of neutrons                   =          31 - 15 = 16

(iii) Magnesium ion (Mg²⁺)

Mg²⁺ ion is formed by the loss of two electrons by Mg atom. Therefore, it has two electrons less than the number of electrons is Mg atom.

Number of electrons                  =          12 - 2 = 10

Number of protons                    =          12

Number of neutrons                   =          (24 - 12_ = 12

2.The number of protons in the nucleus of an atom of mass number 97 is 41. What will be the number of neutrons in its isotope of mass number 99 ?

Sol.  The atomic number of isotopes is same. Therefore, the number of protons in both the atoms is same.

Mass number                           =          Number of protons + Number of neutrons

Number of neutrons              =          Mass number - Number of protons

                                                 =          99 - 41 = 58

3.Give number of protons and neutrons in 

Sol. From the given symbol it is clear that the atomic number of uranium is 92 and its mass number is 238.

Now, number of protons            =          Atomic number = 92

Number of neutrons                 =          Mass number - Atomic number

                                                            =          238 - 92 = 146

4.Calculate the atomic number of an element whose mass number is 31 and number of neutrons is 13. What is the symbol of the element ?

Sol. We know that, mass number = Number of protons + Number of neutrons

But number of protons is equal to the atomic number.

Mass number                     =          Atomic number + number of neutron

or    Atomic number                   =          Mass number - number of neutron = 31 - 16 = 15

The element with atomic number 15 is phosphorus which has symbol P.

5.If bromine atom is available in the form of, say two isotopes, calculate the average atomic mass of bromine atom.

Sol.  % of Br (79) = 49.7 ; % of Br (81) = 50.3

Atomic mass of Br =   

                                 =          80.0

Thus, atomic mass of bromine is 80.0

6.An isotope has atomic number 17 and mass number 37. What is the arrangement of electrons in the shells of this isotope ? State nuclear composition of this isotope.

Sol. Number of electrons      =          Atomic number   = 17

Number of protons        =             Atomic number        = 17

Number of neutrons     =          Mass number - Atomic number = 37 - 17 = 20

Electronic configuration of the isotope is         K          L          M

                                                                             2          8          7         

Nucleus of the isotope contains  17 protons  and 20 neutrons.

7.An element has 2 electrons is the M-shell. What is the electronic configuration of the element and what is its atomic number ?

Sol. The 2 electrons in M-shell indicates that the K and L - shell must be full. K - shell can accommodate a maximum of 2 electrons while L-shell can accommodate a maximum of 8 electrons. Thus, the electronic configuration of the element may be written as :

K          L          M

2           8          2

The total number of electrons is an atom of the element is 2 + 8 + 2 = 12

Therefore, atomic number of element is 12.

8.How will you find the valency of chlorine, sulphur and magnesium ?

Sol. (i) The atomic number of Cl is 17. Its electronic configuration is

K                L          M

2                8          7

Cl has 7 electrons in the valence shell. It needs one more electron to complete its octet. Hence, its valency is 1.

(ii) The atomic number of S is 16. Its electronic configuration is

K                L          M

2                8          6

S has 6 electrons is the valence shell. It requires two more electrons to complete its octet. Hence, its valency is 2.

(iii) The atomic number of Mg is 12. Its electronic configuration is

 K                L          M

 2                8          2

Mg has only 2 electrons in the valence shell. By losing these 2 electrons it can attain octet of electrons in its outer most shell. Hence, it valency is 2.

"Structure Of Atom" 

⇒ First recorded evidence that atoms existed.

⇒ Using his theory, Dalton rationalized the various laws of chemical combination

Dalton's theory was based on the premise that the atoms of different elements could be distinguished by differences in their weights.

⇒ Limitations

o The indivisibility of an atom was proved wrong , for, an atom can be further subdivided into protons, neutrons and electrons.

o The atoms of same element are similar in all respects , but isotopes of same element have different mass.

Dalton's theory was based on the premise that the atoms of different elements could be distinguished by differences in their weights.

2. J J Thomson Experiments:

⇒ Discovered electrons in 1897.

⇒ Showed us that the atom can be split into even smaller parts.

His discovery was the first step towards a detailed model of the atom .

⇒ An atom is a uniform sphere of positive charges (due to presence of protons) as well as negative charges (due to presence of electrons).

⇒ Atom as a whole is electrically neutral because the negative and positive charges are equal in magnitude.

⇒ An electron is a negatively charged component of an atom which exists outside the nucleus. Each electron carries one unit of negative charge and has a very small mass as compared with that of a neutron or proton.

JJ Thomson used cathode ray tubes to demonstrate that the cathode ray responds to both magnetic and electric fields.

Since the ray was attracted to a positive electric plate placed over the cathode ray tube (beam deflected toward the positive plate) he determined that the ray must be composed of negatively charged particles.

He called these negative particles "electrons."

Limitation: Model failed to explain how protons and electrons were arranged in atom so close to each other.

Eugene Goldstein:

⇒ E. Goldstein in 1886 discovered the presence of new radiations in a gas discharge and called them canal rays. These rays were positively charged radiations which ultimately led to the discovery of another sub-atomic particle.

⇒ Used a Cathode Ray Tube to study "canal rays" which had electrical and magnetic properties opposite of an electron

⇒ Canal Rays: The positively charged radiation produced in the discharge tube at low pressure and high voltage are called canal rays.

Protons:

The canal rays have positively charged sub-atomic, particles known as protons (p).

Q.1 What was the model of an atom proposed by Thomson? Q.2 What is the nature of charge on electrons?

Q.3 What are canal rays ? 

Q.4 State the nature of the constituents of canal rays.

3. Rutherford’s Scattering Experiments:

Experiment: Rutherford took a thin gold foil and made alpha particles , [ He2+ ] positively charged Helium fall on it.

⇒ Limitation: In Rutherford’s atomic model , Nucleus & electrons are held together by electrostatic force of attraction which would lead to the fusion between them. This does not happen in the atom.

Atomic radius ~ 100 pm = 1 x 10^-10 m

Nuclear radius ~ 5 x 10^-3 pm = 5 x 10-¹⁵ m

⇒ In 1932, James Chadwick proved that the atomic nucleus contained a neutral particle which had been proposed more than a decade earlier by Ernest Rutherford officially discovered the neutron in 1932,

⇒ Chadwick received the Nobel Prize in 1935.

A neutron is a subatomic particle contained in the atomic nucleus. It has no net electric charge, unlike the proton's positive electric charge.

Q.1 Who discovered the nucleus of the atom?

Q.2 What is the charge on alpha particles ?

Q.3 Which observation of Rutherford’s scattering experiment established the presence large empty space in atom?

Q.4 What is the nature of charge on nucleus of atom? Q.5 Who discovered neutron ?

4. Sub Atomic Particles:

Protons & Neutrons collectively are known as NUCLEONS.

Q.1 Why is the relative mass of an electron is taken as negligible ?

Q.2 Give the actual masses of electron & proton in kg?

Q.3 What are nucleons?

5.Atomic Number & Mass Number:

“Atomic number of an element is defined as the number of unit positive charges on the nucleus (nuclear charge) of the atom of that element or as the number of protons present in the nucleus.”

Atomic number, Z = Number of unit positive charge on the nucleus

= Total number of unit positive charges carried by all protons present in the nucleus.

= Number of protons in the nucleus (p)

= Number of electrons revolving in the orbits (e)

Eg :- Hydrogen – Atomic number = 1 (1 proton)

        Helium - Atomic number = 2 (2 protons)

Mass number[ A] : It is defined as the sum of the number of protons & neutrons present in the nucleus of an atom.

Mass Number = Mass of protons + Mass of neutrons

Eg :- Carbon – Mass number = 12 (6 protons + 6 neutrons) Mass = 12u

         Aluminium – Mass number = 27 (13 protons + 14 neutrons) Mass = 27u

Q.1 What happens when an electron jumps from lower to higher energy level?

Q.2 Which energy shell is nearest to the nucleus of an atom?

Q.3 Which energy shell has higher energy L or N ?

7.Electronic configuration & Valency:

Bohr and Bury Scheme - Important Rules

The outermost shell of an atom cannot accommodate more than 8 electrons, even if it has a capacity to accommodate more electrons. This is a very important rule and is also called the OCTET RULE. The presence of 8 electrons in the outermost shell makes the atom very stable.

Q.1 An atoms has atomic number 13. What would be its configuration.

Q.2 What is octet rule?

Q.3 How many electrons M shell can accommodate?

Q.4 If an atom has complete K and L shell, what would be its atomic number?

8. Isotopes & Isobars:

You are expected to know…………

⇒ The scientists who discovered subatomic particles.

⇒ Rutherford established the existence of nucleus.

⇒ Difference between Atomic number and Mass number

⇒ Electronic configuration & its relation with Valency.

⇒ Difference between Isotope and Isobar.

Chapter Name: Structure of atom Top concepts

1. Ionisation of gases in the discharge tube proved that atoms have subatomic particles

2. Summary of characteristics of electrons, protons and neutrons

3. Thomson model of atom : An atom is a uniform sphere of positive charges (due to presence of protons) as well as negative charges (due to presence of electrons). Atom as a whole is electrically neutral because the negative and positive charges are equal in magnitude

4. Limitations of Thomson model of atom: Model failed to explain how protons and electrons were arranged in atom so close to each other

5. α -particles are charged particles having two units of positive charge and 4 units of mass, i.e. they are doubly charged helium ions(He²⁺)

6. Observations predicted from α -particle scattering experiment by Rutherford based on Thomson model of atom:

Rutherford expected that if the model proposed earlier by J.J Thomson, according t0 which there is uniform distribution of positive and negative charge, was correct then α-particle particles striking the gold atoms would be uniformly deflected which was not the case.   Since the -particles were much heavier than the protons, he did not expect to see large deflections

7. Selection of gold metal for Rutherford's  α -particle scattering experiment:

Since α -particles are easily absorbed by objects; he wanted to ensure that alpha particles pass through foil without getting absorbed.

If α -particles were absorbed in matter then they would not be able to give any useful information about insight of atom.

Only if α-particles were deflected then only it can give useful information about insight of atom. Gold is easily malleable and can be beaten into very thin sheets.

8. Observations made by Rutherford from  α-particle scattering experiment:

1. Most of the α-particles passed straight through gold foil without suffering any deflection from their original path

2. Some of the α -particles were deflected by the foil by small angles

3. One out of every 12000 particles appeared to rebound

9. Conclusions from Rutherford’s model of an atom

1. Most of the space between inside atoms is empty; hence it allows the α particles to pass straight through it without any deflection

2. Very few particles were deflected from their path which suggests that the positive charge of the atom occupies very little space

3. The total volume occupied by a nucleus is very small compared to the total volume of the atom, as very few  α particles are reflected by 180o and all the positive charge and mass of the gold atom were concentrated in a very small volume within atom.

10. Rutherford’s model of an atom

1. There is positively charged centre in an atom called the nucleus and the entire mass of atom resides in the nucleus

2. Electrons revolve around the nucleus in well defined orbits

3. Size of nucleus is very small as compared to size of atom 

11. Defects in Rutherford model of atom:

1. Rutherford had proposed that electrons move around a positively charged nucleus at very high speed in circular orbi To remain in a circular orbit electron would have to be accelerated centripetally (tending to move toward a center). But according to electromagnetic theory if charged body (electron) is accelerated around another charged body (nucleus) then there would be continuous radiation of the moving body (i.e. electron). This loss of energy would slow down speed of electron and eventually electron would fall into nucleus. But Rutherford’s model could not explain such a collapse

2. Rutherford had proposed that electrons revolve around the nucleus in fixed orbi He did not specify the number of electrons in each orbit

12. Postulates put forward by Bohr regarding model of atom:

1. Electrons revolve around the nucleus in a limited number of orbits called discrete orbits of electrons or also called as permissible orbits

2. While revolving in discrete orbits the electrons does not radiate energy i.e. energy of an electron remains constant so long as it stays in a given orbi Electrons present in different orbits have different energies

3. When an electron jumps from lower energy level to higher energy level some energy is absorbed, while energy is released when electron jumps from higher energy to lower one Orbits or shells are represented by the letters K, L, M, N… or the numbers, n=1, 2, 3, 4…

13. Bohr-Bury scheme for distribution of electrons in different orbits

1. Maximum number of electrons that can be accommodated in a shell is given by 2n2 ,where n is the shell number i.e. first shell can accommodate 2 electrons, second shell can accommodate 8 electrons, third shell can accommodate 18 electrons and so on

2. Outermost orbit of an atom can accommodate a maximum number of 8 electrons

3. Electrons are not accommodated in a given shell, unless the inner shells are filled i.e. the shells are filled in a step-wise manner

14. Combining capacity of an atom is called its valency

15. Outermost shell of an atom is called valence shell

16. Electrons present in valence shell is called valence electrons

17. If the outermost shell of an atom is completely, its valency is 0

18. Valency of elements having 1-4 electrons in outermost shell are generally determined by the rule: Valency = Number of electrons in outermost shell

19. Valency of elements having number of electrons in outermost shell close to 8 is determined by the formula: Valency=8- Number of electrons in outermost shell

20.Significance of valence electrons:

1. Valence electrons are responsible for chemical changes

2. Elements having same number of valence electrons in their atoms possess similar chemical properties because chemical properties of an element are determined by the number of valence electrons in an atom

3. Elements having different number of valence electrons in their atoms possess different chemical properties

21. Atomic number(Z) is defined as the total number of protons present in nucleus of an atom

22.Protons and neutrons together are called nucleons

23.Mass number (A) is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom.

24. Isotopes:

♦Isotopes are the atoms of same element having same atomic number but different mass number

♦Isotopes have similar chemical properties because they have same number of valence electrons

♦Isotopes have different physical properties like boiling point, melting point etc because they have different mass number

♦Fractional atomic mass of elements are due fact that all elements have isotope

Applications of isotopes:

1. Uranium isotope( 235U) is used in nuclear reaction 92

2. Cobalt isotope ( 60U ) is used to remove brain tumours and in their treatment 27

3. Isotope of sodium ( 24Na)has been used to diagnose restricted circulation of blood 11

25. Isobars are the atoms of different element with different atomic number but same mass number

CHAPTER 2: STRUCTURE OF ATOM IMPORTANT QUESTIONS:-

ONE MARK QUESTIONS

Q.1. Neutrons can be found in all atomic nuclei except in one case. Which is this atomic nucleus and what does it consists of?

Ans. Hydrogen atom. It consists of only one proton.

Q.2. 14. Mention the draw backs of Rutherford’s atomic model.

Ans.. 1. It could not explain the stability of an atom.

2. It could not explain the line spectrum of H- atom.

Q.3. What are the values of n and l for 2p orbital?

Ans. n=2 and l= 1

Q.4. Which of the following orbitals are not possible? 1p, 2s, 3f and 4d

Ans. 1p and 3f are not possible.

Q.5. Write the electronic configuration of the element having atomic number 24.

Ans. 1s2 2s2 2p6 3s2 3p6 3d5 4s1

Q.6. Which orbital is non-directional ?

Ans. s- orbital

Q.7. Does the nature of cathode ray depend on the nature of gas in the discharge tube or the electrode material ?

Ans. No. Neither nature of the gas nor electrode material

Q.8. What is the value of speed of light in vacuum?

Ans. 3.0x108m/s

Q.9. Write down the quantum numbers n and l for the following orbitals

1. 2p b. 3d c. 5f

Ans. a. n=2, l= 1 b. n= 3, l=2 c. n= 5, l=3

Q.10. Write the complete symbol for the atom with the given atomic number (Z) and mass number(A). (a) Z = 17, A = 35 (b) Z = 92 , A = 233 35 233

Ans. (a) 17Cl (b) 92U

Q.1. Calculate wave number of yellow radiations having wavelength of 5800 A0.

Ans. Wave number = 1/ wavelength

Wavelength = 5800 A0 = 5800 x 10-10 m

Wave number = 1/5800 x 10-10 m

= 1.72 x 106 m-1

Q.2. Using s,p,d and f notation, describe the orbital with the following quantum numbers-

(a) n=1,l=0 (b) n=3, l=1 (c) n=4, l=2 (d) n=4, l=3

Ans. (a) 1s (b) 3p (c)4d (d) 4f

Q.3. (a)What is the lowest value of n that allows g orbitals to exist?

(b)An electron is in one of the 3d orbitals, Give the possible values of n, l and ml for this electron.

Ans.(a) minimum value of n= 5

(b)n=3, l=2, ml = -2, -1, 0, +1, +2

Q.4. Write down the quantum numbers n and l for the following orbitals

1. 2p b. 3d c. 5f

Ans. a. n=2, l= 1 b. n= 3, l=2 c. n= 5, l=3

Q.5. How many electrons in an atom have the following quantum numbers?

1. n=4, ms= -1/2 b. n =3 , l=o

Ans. (a) 16 electrons (b) 2 electrons.

Q.6. Calculate the total number of angular nodes and radial nodes present in 3p orbitals.

Ans. For 3p orbitals, n=3, l= 1 Number of angular nodes = l= 1 

Number of radial nodes = n-l-1= 3-1-1= 1

Q.7. (i) The energy associated with the first orbit in the hydrogen atom is -2.18 x 10-18J/atom.

(ii) What is the energy associated with the fifth orbit

Ans. (i) En = -2.18 x 10-18/ n2

(ii) E5 = -2.18 x 10-18/ 52 = -8.72 x 10-20 J

Q.8. Explain , giving reasons, which of the following sets of quantum numbers are not possible.

(a) n=0, l=0; ml = 0, ms= + ½               (c) n=1, l=0; ml = 0, ms= - ½

(b) n=1, l=1; ml =- 0, ms= + ½             (d) n=2, l=1; ml = 0, ms= + ½

Ans. (a) Not possible because n≠ 0 

(b) Possible

(c) Not possible because when n=1, l≠1

(d) Possible

Q.9. Write 3 points of differences between electromagnetic waves and matterwaves.

Q.10. Write the 3 points of difference between orbit and orbital.

Q.1. (a) What are nodes?

(b) Write the total number of nodes for a given value of ‘n’.

(c) The total number of nodes for 3s orbital is ------------------

Ans. (a)The region where probability density function reduces to zero.

(b)Total no. of nodes = (n-1).

(c)No. of nodes = 2

Q.2. Calculate the energy of one mole of photon of radiation whose frequency is 4X1012 Hz.

Q.3. State de-Broglie concept of dual nature of matter. How do dual nature of electron verified?

Ans. Just as light has dual nature, every material particle in motion has dual nature (particle nature and wave nature). The wave nature has been verified by Davisson and Germer’sexperiment whereas particle nature by scintillation experiment.

Q.4. State (a)Hund’s Rule of maximum Multiplicity

(b) Aufbau Principle

(c) n+l rule

Ans.(a) Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.

(b)In the ground state of the atoms, the orbitals are filled inorder of their increasing energies.

(c) Orbitals with lower value of (n+l) have lower energy.If two orbitals have the same value of (n+l) then orbital with lower value of n will have lower energy. It is also known as Bohr Burry’s rule

Q.5. Write down the quantum numbers n and l for the following orbital

a.2p          b. 3d          c. 5f

Ans. a. n=2, l= 1         b. n= 3, l=2        c. n= 5, l=3

Q.6. An atom of an element contains 29 electrons and 35 neutrons. Deduce

(i)the number of protons and

(ii) the electronic configuration of the element

(iii) Identify the element .

Ans.(i)For an atom to be neutral, the number of protons is equal to the number of electrons. ∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is 1s22s2 2p6 3s2 3p64s1 3d10

(iii) Copper

Q.7. What are the Characteristics of Canal Rays?

Ans. i. Positively charged particles (Canal Rays) depend up on the nature of the gas present in the cathode-ray tube.

ii.The charge to mass ratio of the particles depend on the gas from which they originate.

iii.Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.

iv.In the presence of electrical or magnetic field, the behavior of positively charged particles is opposite to that observed for cathode rays.

Q.8. What is the significance of

(a) principal quantum number(n);

(b) azimuthal quantum number (l);

(c) magnetic quantum number (ml);

(d) spin quantum number (ms)?

Ans. a) It determines the size and energy of the orbital.

b) It determines the three dimensional shape of the

c) It gives the spatial orientation of the orbital.

d) It refers to orientation of the spin of the electron

Q.9. Write the electronic configuration of the following elements: (a) Cr (Z=24) (b) Cu (Z=29) (c) Ca (Z=20).

Ans.